\newproblem{lay:3_3_7}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.3.7}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Determine the values of the parameter $s$ for which the system below has a unique solution.
	\begin{center}
		$6sx_1+4x_2=5$\\
		$9x_1+2sx_2=-2$
	\end{center}
}{
   % Solution
	Applying Cramer's rule
	\begin{center}
		$x_1=\frac{\left|\begin{array}{rr} 5 & 4 \\ -2 & 2s\end{array}\right|}{\left|\begin{array}{rr} 6s & 4 \\ 9 & 2s\end{array}\right|}=
		     \frac{5\cdot 2s - 4\cdot (-2)}{6s\cdot 2s - 4\cdot 9}=\frac{10s+8}{12s^2-36}=\frac{10(s+\frac{8}{10})}{12(s+\sqrt{3})(s-\sqrt{3})}
				=\frac{5(s+\frac{8}{10})}{6(s+\sqrt{3})(s-\sqrt{3})}$ \\
		$x_2=\frac{\left|\begin{array}{rr} 6s & 5 \\ 9 & -2\end{array}\right|}{\left|\begin{array}{rr} 6s & 4 \\ 9 & 2s\end{array}\right|}=
		     \frac{6s\cdot (-2) - 5\cdot 9}{6s\cdot 2s - 4\cdot 9}=-\frac{12s+45}{12s^2-36}=-\frac{12(s+\frac{45}{12})}{12(s+\sqrt{3})(s-\sqrt{3})}
				=-\frac{s+\frac{45}{12}}{(s+\sqrt{3})(s-\sqrt{3})}$ \\
	\end{center}
	
	This equation system has a unique solution if the denominator of the fractions above do not vanish, that is, $s\neq \pm\sqrt{3}$.
}
\useproblem{lay:3_3_7}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
